Dave Dickey Forum Moderator 324 Posts
Re: Is it possible to directly select the gear reducer based on motor input power?8 July 2014 at 2:07pm
Don't try to make this problem more difficult than it is. The answer to your basic question is yes. It is possible, even recommended, to directly select the gear reducer based on motor input power. The gear reducer selection should be made on the basis of the motor power and the reduced output speed. The maximum torque will always be less than the power the motor can deliver at the reduced output speed, otherwise the motor will over load. If the existing process conditions do not overload a 2.2 kW motor with an output speed or 36 rpm (0.6 1/s), then the gear reducer should be sized for a 2.2 kW motor. 2.2 kW = 2,000 W = 2,000 Nm/s. Torque = power / speed = 2,000 Nm / (0.6 /s) = 3,333 Nm.
I have with me an existing agitator, details of which are as below:Motor = 2.2KW/ 3 HP, 1440 rpm gear reducer = ratio 40/1, agitator shaft rpm = 36, agitator shaft dimension length = 2500mm round bar, SS 304 blades = 5 nos, 100 mm wide, 5 mm thick, diameter = 2200 mm, at a distance of 400 mm each, SS 304 weight of the agitator shaft and impeller assembly = approx 177 lbs.Purpose of agitation: Alum tank agitator density ~ 1.3 - 1.4 Now I need the one more exact assembly to increase the capacity of my plant, however I have no clue how to derive the absorbed power at the agitator shaft or the torque requirement to drive the agitator shaft. I have basic knowledge of gear reducer selection for which I need to calculate the absorbed power. Can someone please suggest how to do that. Also, is it possible to directly select the gear reducer based on the motor input KW?
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